LeetCode 1229: Meeting Scheduler Solution

Master LeetCode problem 1229 (Meeting Scheduler), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1229. Meeting Scheduler

Medium

Array Two Pointers Sorting

Problem Explanation

Explanation

The problem asks us to find a common meeting time slot of a specified duration between two schedules of meetings. We need to find the earliest possible time slot that satisfies the duration requirement.

To solve this problem, we can follow these steps:

Merge the schedules of both persons.
Sort the merged schedule based on the start time.
Iterate through the sorted schedule to find a time slot that satisfies the duration requirement.

Time Complexity: O(n log n), where n is the total number of slots in both schedules. Space Complexity: O(n), where n is the total number of slots in both schedules.

Solution Code

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;

class Solution {
    public List<Integer> minAvailableDuration(int[][] slots1, int[][] slots2, int duration) {
        List<int[]> merged = new ArrayList<>();
        for (int[] slot : slots1) {
            merged.add(slot);
        }
        for (int[] slot : slots2) {
            merged.add(slot);
        }
        
        Collections.sort(merged, (a, b) -> a[0] - b[0]);
        
        for (int i = 0; i < merged.size() - 1; i++) {
            int[] current = merged.get(i);
            int[] next = merged.get(i + 1);
            if (next[0] - current[1] >= duration) {
                return List.of(current[1], current[1] + duration);
            }
        }
        
        return new ArrayList<>();
    }
}

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